Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $n = \dfrac{5k - 20}{-5k^2 - 45k + 50} \div \dfrac{-2k + 16}{k^2 + 2k - 80} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{5k - 20}{-5k^2 - 45k + 50} \times \dfrac{k^2 + 2k - 80}{-2k + 16} $ First factor out any common factors. $n = \dfrac{5(k - 4)}{-5(k^2 + 9k - 10)} \times \dfrac{k^2 + 2k - 80}{-2(k - 8)} $ Then factor the quadratic expressions. $n = \dfrac {5(k - 4)} {-5(k + 10)(k - 1)} \times \dfrac {(k + 10)(k - 8)} {-2(k - 8)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {5(k - 4) \times (k + 10)(k - 8) } { -5(k + 10)(k - 1) \times -2(k - 8)} $ $n = \dfrac {5(k + 10)(k - 8)(k - 4)} {10(k + 10)(k - 1)(k - 8)} $ Notice that $(k + 10)$ and $(k - 8)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {5\cancel{(k + 10)}(k - 8)(k - 4)} {10\cancel{(k + 10)}(k - 1)(k - 8)} $ We are dividing by $k + 10$ , so $k + 10 \neq 0$ Therefore, $k \neq -10$ $n = \dfrac {5\cancel{(k + 10)}\cancel{(k - 8)}(k - 4)} {10\cancel{(k + 10)}(k - 1)\cancel{(k - 8)}} $ We are dividing by $k - 8$ , so $k - 8 \neq 0$ Therefore, $k \neq 8$ $n = \dfrac {5(k - 4)} {10(k - 1)} $ $ n = \dfrac{k - 4}{2(k - 1)}; k \neq -10; k \neq 8 $